If the half - life of the first order reaction is 50 s, what be the value of its rate constant ?

Options

A. `1.38 xx10^(-2) s^(-1)`
B. `25 s^(-1)`
C. `34.66s^(-1)`
D. `1.38xx10^(-4)s^(-1)`

Answer

To find the rate constant (k) for a first-order reaction given its half-life (t₁/₂), we can use the formula that relates the two:

### Step-by-Step Solution:

1. **Understand the relationship**: For a first-order reaction, the half-life (t₁/₂) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Substitute the given half-life**: We are given that the half-life \( t_{1/2} \) is 50 seconds. Substitute this value into the formula:
\[ 50 = \frac{0.693}{k} \]

3. **Rearrange the equation to solve for k**: To find \( k \), we can rearrange the equation:
\[ k = \frac{0.693}{50} \]

4. **Calculate the value of k**: Now, perform the calculation:
\[ k = \frac{0.693}{50} = 0.01386 \, \text{s}^{-1} \]

5. **Final answer**: Therefore, the rate constant \( k \) for the reaction is approximately:
\[ k \approx 0.0139 \, \text{s}^{-1} \]